University of Manchester Department of Computer Science CS3282 Digital Communications ’05-’06 Section 5: Detection of Binary Signals in AWGN Mar'06 CS3282 Sectn 5 1
•Consider transmission of single bit using binary signalling. •At receiver in absence of noise, let symbols be s 1 (t) and s 0 (t) . •If received signal is affected by AWGN n(t), we receive: r(t) = s i (t) + n(t) : i=1 or 0 • Assume n(t) has ‘2-sided’ PSD N 0 /2 Watts/Hz. • Power & bandwidth of noise n(t) is not specified, just PSD. • Simple example is 'unipolar' signalling with 'rectangular' pulse shape. • Alternative is to receive 'bipolar' signalling with a rect pulse shape. • For unipolar signalling, with AWGN, received signal r(t) will be as shown for ' 1 ' & ' 0 ’: Mar'06 CS3282 Sectn 5 2
Bipolar signalling with AWGN s 1 (t)+n(t) s 0 (t)+n(t) +V +V t t T Mar'06 CS3282 Sectn 5 3
• Restrict analysis to single bit to begin with. • For unipolar signalling, strategy for detecting 1 or 0 would be to sample r(t) at t=T/2 , & compare it with a 'threshold' +V/2 . • If r(T/2) > +V/2 , likely s 1 (t) otherwise s 0 (t) . • For rect pulses, no matter whether we sample in middle, at beginning or at end . • Wherever we sample, risk that noise n(t) will cause s 1 (t) to be mistaken for s 0 (t) or vice-versa. • Choice of threshold half way between zero & +V is good when probabilities of s 0 (t) & s 1 (t) are known to be equal, i.e. 0.5. • This is commonly the case. Mar'06 CS3282 Sectn 5 4
Exercise 5.1: For 0 & 1 volt binary symbols probabilities are 0.1 & 0.9 rather than 0.5 & 0.5. Variance of noise at detector is 1/16 . What is bit-error prob P B if threshold γ = 0.5? Can we reduce P B by choosing a different value of γ ? Solution: σ = ¼ = 0.25. With γ = 0.5, P B = 0.1 Q (0.5/ σ ) + 0.9 Q (0.5/ σ ) = Q(0.5/ σ ) as usual. Now d Q(z) /dz = − (1/ √ (2 π )) exp(-z 2 / 2) and P B = prob(0)*Q( γ / σ ) + prob(1)*Q((1- γ )/ σ ) dP B /d γ = (-1/( σ√ (2 π )) [prob(0) exp(- γ 2 /(2 σ 2 ) − prob(1)exp(-(1- γ ) 2 /(2 σ 2 ) ) ] dP B /d γ =0 if prob(0) exp(- γ 2 /(2 σ 2 ) = prob(1)exp(-(1- γ ) 2 /(2 σ 2 ) ) i.e. log e (prob(0)) - γ 2 /(2 σ 2 ) = log e (prob(1)) - (1- γ ) 2 /(2 σ 2 ) ∴ γ best = 0.5 - σ 2 log e ( prob(1 volt) / prob(0) ) Mar'06 CS3282 Sectn 5 5
Exercise 5.1 (continued) So we can find a value of γ that gives a lower value of P B as derived on previous slide. Its value is: γ best = 0.5 - σ 2 log e ( prob(1) / prob(0) ) = 0.5 – [log e (9)]/16 = 0.36 P B = prob(0)*Q( γ / σ ) + prob(1)*Q( (1- γ ) / σ ) = 0.1 Q(0.36/0.25) + 0.9 Q(0.64/0.25) = 0.1Q(1.44) + 0.9Q(2.56) = ... Can use trial & error with Q(z) curve, or a simple MATLAB program to check this. Remember that Q(z) = 0.5*erfc( z / ( √ 2) ) Mar'06 CS3282 Sectn 5 6
% Investigate effect of choice of threshold on unipolar signalling % when probabilities of ones and zeros are not equal. prob1=0.9; prob0=0.1; % Voltages are 1 and 0 nstdev = 1/4 ; % Standard dev of AWGN pemin = 1; % Becomes minimum bit-error probability bestgamma=0; % Becomes best thrshold between 0 & 1 for i=10:90 gamma=i/100; pe = prob0*0.5*erfc((gamma/nstdev)/1.414); pe = pe + prob1*0.5*erfc(((1-gamma)/nstdev)/1.414); disp(sprintf('gamma=%f pe=%g',gamma,pe)); if pe<pemin pemin = pe; bestgamma=gamma; end end; disp(sprintf('bestgamma=%f',bestgamma)); disp(sprintf('pemin=%g',pemin)); Mar'06 CS3282 Sectn 5 7
Improved detectn process for unipolar rect symbols s 1 (t) or s 0 (t) : z(t) r(t) Decide z(T) z(T)>V thres Sample at t=T Averager = a i (t)+n 0 (t) = s i (t)+n(t) z(T)<V thres Response of averager to s i (t) is a i (t) , & response to n(t) is n 0 (t) . Averager (or smoother) could ideally produce an output t 1 ∫ = τ τ z ( t ) ( ) r ( ) d T 0 Mar'06 CS3282 Sectn 5 8
t ∫ τ τ To generate r ( ) d 0 -z(t) r(t) - + Dumper Integrator • Force integrator output to start at zero volts at t=0. (A switch dumps charge on capacitor.) • ’Integrate and dump' circuit. • For unipolar signalling with rect pulse, response a 0 (t) of averager to s 0 (t) without noise would be zero. • Response a 1 (t) to s 1 (t) would be as shown on next slide. • Starts at zero & reaches +V at t=T . • Remains at +V until we apply the charge dumper again. Mar'06 CS3282 Sectn 5 9
a 1 (t) +V t T •Instead of clean a 1 (t) , averager produces z(t) = a 1 (t) + n 0 (t) where n 0 (t) is due to n(t) . •Averager reduces noise. •Hence chances of noise causing wrong decision reduced. •Sample at t=T since averaging must be allowed to finish. •After z(t) sampled, charge dumping sets averager output to zero, •Circuit then able to receive another pulse. Mar'06 CS3282 Sectn 5 10
Matched filter' method for detecting rectangular pulse shape. •Use a filter rather than an averager. •Denote filter response to r(t) as ψ (t) ( ψ is PSI) • a i (t) & n 0 (t) now different from what they were with averager. Decide ψ (T) Sample at ψ (T) >V thres Filter ψ (t) r(t) t=T ψ (T) <V thres • Low-pass filter is sort of averager, but there are differences. • Let frequency-response of filter be H( f ) which is FT of h(t) . • Assume that the impulse-response is: Mar'06 CS3282 Sectn 5 11
Filter’s impulse resp: ≤ ≤ 1 / T : 0 t T = h ( t ) 0 : otherwise Reason for this impulse-response will be made clear later. In response to r(t), filter's output is: ∞ ∫ ψ = τ − τ τ ( t ) h ( ) r ( t ) d − ∞ T ∫ ψ = − τ τ ( t ) ( 1 / T ) r ( t ) d In this case: 0 T ∫ ψ = − τ τ ( T ) ( 1 / T ) r ( T ) d and at t=T: 0 Substituting t=T- τ which means that d τ = -dt, we obtain: 0 T ∫ ∫ ψ = − = ( T ) ( 1 / T ) r ( t ) dt ( 1 / T ) r ( t ) dt T 0 Mar'06 CS3282 Sectn 5 12
•Identical to value of z(T) we obtained using the averager. •Another way of generating z(T) for a single pulse. •Uses a filter rather than an averaging circuit. •Note ψ (t) equals z(t) only at t=T, but this is only point we need. •Response of filter to s 1 (t) is as follows: a 1 (t) +V t 2T T Filter output decays to zero at t=2T without 'charge dumper'. Mar'06 CS3282 Sectn 5 13
Estimating bit-error rate when ‘ I & D’ or ‘MF’ is employed. • Bit-error rate will be the same for both approaches. • Analyze matched filter approach: • Consider 2-sided PSD of n 0 (t) , where n 0 (t) is filter's response to n(t) . PSD 0 ( ƒ ) = PSD(f) |H(f)| 2 where PSD( ƒ ) is the ‘2-sided’ PSD of noise n(t) . • Power of n 0 (t) is therefore: ∞ ∞ ∫ ∫ = 2 PSD ( f ) df PSD ( f ) | H (( f )) | df 0 − ∞ − ∞ ∞ ∫ 2 = = 0 . 5 N H (( f )) df 0 . 5 N / T 0 0 − ∞ since PSD( f )=0.5 N 0 constant for all f , and, by Parseval’s Theorem, ∞ ∞ T ∫ 2 ∫ 2 ∫ = = = 2 H (( f )) df h ( t ) dt ( 1 / T ) dt 1 / T − ∞ − ∞ 0 •When AWGN applied to filter, output is coloured Gaussian noise. Mar'06 CS3282 Sectn 5 14
To summarise: • If white Gaussian noise n(t) with 2-sided PSD N 0 /2 Watts/Hz were applied to a filter with impulse-response: ≤ ≤ 1 / T : 0 t T = h ( t ) 0 : otherwise output would be Gaussian noise n 0 (t) of power 0.5N 0 /T Watts. ∴ if r(t) = s i (t) + n(t) , with i=1 or 0 , applied to same filter, output is: ψ (t) = a i (t)+n 0 (t) where a i (t) is filter's response to s i (t) for i = 1 or 0. • If we sample ψ (t) at t=T , ψ (T) = a i (T)+n 0 (T) is equal to z(T) as obtained by sampling output z(t) from an averaging circuit at t=T. Mar'06 CS3282 Sectn 5 15
• In case of unipolar signalling with + V & zero valued rect pulses, we know that a i (T) is + V or zero , but what can we say about n 0 (T) ? • It is a voltage obtained by sampling a random signal which may cause a bit-error depending on how large it is. • Cannot say what this voltage will be exactly so we cannot say definitely that it will or will not cause an error. • Can estimate probability of it being large enough to cause an error • Use properties of statistical process that describes production of this voltage ; i.e. a statistical process which is Gaussian with zero mean & variance 0.5N 0 /T . • Subject to certain conditions, if average value of a random voltage signal is zero, its power equals the variance ( σ 2 ) of a statistical process that describes it. Mar'06 CS3282 Sectn 5 16
• Effect of “I&D” or MF is to produce a i (T) + n 0 (T) where n 0 (T) is random variable of variance 0.5N 0 /T . • If a i (T) is +V & a 0 (T)=0 , & we take threshold at +V/2 , probability of bit-error is: P b = ' prob(n 0 (T) < − V/2) when s 1 (t) received' + ' prob(n 0 (T) > V/2) when s 0 (t) is received'. • No correlation between noise & bit-stream. Therefore P b = prob(n 0 (T) < − V/2) * prob ( transmitting 1) + ' prob(n 0 (T) > V/2)* prob (transmitting 0) • If '1’ & '0' equally likely, P b = 0.5 [ prob(n 0 (T)<-V/2) + prob(n 0 (T)>+V/2) ] . •For AWGN with mean=0: prob( n 0 (T)<-V/2 ) = prob( n 0 (T)>+V/2 ) • Therefore for unipolar rect signalling, P b = prob ( n 0 (T) > +V/2 ) Mar'06 CS3282 Sectn 5 17
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